what's wrong?

[questions]

# include <stdio.h>
# include <math.h>
int main()
{ long int x,y;

printf("enter an integer\n");
scanf("%d",&x);

y=x%pow(10,3);

printf("the result is %d",y);

return 0;}

The compiler tell me there is something wrong with the "pow",but I
don't know what's the wrong?

[Kai-Uwe Bux]

questions wrote:

> # include <stdio.h>
> # include <math.h>
> int main()
> { long int x,y;
>
> printf("enter an integer\n");
> scanf("%d",&x);
>
> y=x%pow(10,3);

[snip]

> The compiler tell me there is something wrong with the "pow",but I
> don't know what's the wrong?

pow() returns a floating point (in this case, I guess a double) and x is a
long int. That is not a valid pair of operand types for the % operator.

Best

Kai-Uwe Bux

[Zeppe]

my compiler says:

testmod.cpp:8: warning: format ‘%d’ expects type ‘int*’, but argument 2
has type ‘long int*’
testmod.cpp:10: error: invalid operands of types ‘long int’ and ‘double’
to binary ‘operator%’
testmod.cpp:12: warning: format ‘%d’ expects type ‘int’, but argument 2
has type ‘long int’

It looks pretty much clear. Listen to your compiler. It's your friend.

Best wishes,

Zeppe

[Juha Nieminen]

questions wrote:
> y=x%pow(10,3);

What's wrong with 1000?

[Kai-Uwe Bux]

Juha Nieminen wrote:

Nit: 1000 does not occur in the above whereas 1000.0 does occur.

Best

Kai-Uwe Bux

[Sjouke Burry]

Another nit : Seeing the % sign suggests that the intent is to use 1000

[Juha Nieminen]

But 1000 would work, whereas 1000.0 wouldn't.

[Kai-Uwe Bux]

Juha Nieminen wrote:

Ah, now I understand your point: you meant to replace pow(10,3) by 1000.

Sorry for the noise.

Kai-Uwe Bux

[questions]

thanks

[questions]

On 10月28日, 下午7时56分, Zeppe

> Zeppe- 隐藏被引用文字 -
>
> - 显示引用的文字 -

thanks

[questions]

thanks